Editing SSC/Dynamics/FreeFall (section)

====Relationship to Kepler's 3<sup>rd</sup> Law====

[[File:CommentButton02.png|right|100px|Note from J. E. Tohline:  I was first made aware of this relationship, as a graduate student, while listening to John Faulkner give a lecture on the free-fall problem to a class of undergraduates at UCSC.]]It is useful to note a relationship between [http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/ Kepler's 3<sup>rd</sup> law] and the free-fall problem, as introduced [[#Falling_from_rest_at_a_finite_distance_.E2.80.A6|here in the context of the motion of a single (massless) particle that falls from rest]] toward a point-like object of mass, <math>~M</math>.  According to Kepler's 3<sup>rd</sup> law, when a massless particle orbits a point-like object of mass, <math>~M</math>, the particle's orbital period, <math>~P_\mathrm{orb}</math>, is related to the semi-major axis, <math>~a_\mathrm{orb}</math>, of its elliptical orbit via the algebraic expression,
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<math>~P^2_\mathrm{orb} = \frac{4\pi a_\mathrm{orb}^3}{GM} \, .</math>
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This relation works for orbits of any eccentricity, 
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<math>~e \equiv \biggl[ 1 - \frac{b_\mathrm{orb}^2}{a_\mathrm{orb}^2} \biggr]^{1/2} \, ,</math>
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where, <math>~b_\mathrm{orb}</math> is the semi-minor axis of the orbit, the extremes being:  <math>~e = 0 ~(b_\mathrm{orb} = a_\mathrm{orb})</math> for a circular orbit, and <math>~e = 1 ~(b_\mathrm{orb} = 0)</math> for a purely radially directed (in-fall) trajectory.  

In the context of our current discussion, it should be clear that a particle that "free-falls" from rest at an initial distance, <math>~r_i</math>, from a point mass object will follow a trajectory synonymous with a Keplerian orbit having eccentricity, <math>~e=1</math>.  The particle's initial position coincides with the apo-center of this orbit and the point mass object is located at the peri-center (as well as at one focus) of the orbit, so the semi-major axis is <math>~a_\mathrm{orb} = r_i/2</math>.  We also recognize that the particle will move from the apo-center to the peri-center of its orbit &#8212; completing its "free-fall" onto the point-mass object &#8212; in a time, <math>~\tau = P_\mathrm{orb}/2</math>.  From Kepler's 3<sup>rd</sup> law, we therefore deduce that,
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<math>~\tau = \frac{1}{2} \biggl[ \frac{4\pi (r_i/2)^3}{GM} \biggr]^{1/2} = \biggl( \frac{\pi r_i^3}{8GM} \biggr)^{1/2} \, ,</math>
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which precisely matches the free-fall time, <math>~\tau_\mathrm{ff}</math>, derived above.